Let $R$ and $S$ be two commutative rings with unity. Prove that $R\times S$ is NOT an integral domain.
This is the best I could think of so far, please give me a push in the right direction and correct me.
It suffices to show that $R\times S$ has zero-divisors
Therefore, let $a$ be an element of $R$ and $b$ be an element of $S$, thus $R\times S$ has elements of the form $(a,b)$.
Consider the elements $A=(a,0)$ and $B=(0,b)$, such that $a$ does not equal zero and $b$ does not equal zero. Then $AB = 0$. Therefore $R\times S$ has zero divisors.